Theorem - Let $\mathcal{A}$ be a complex Banach algebra with identity element. The spectrum of each $a \in \mathcal{A}$ is a non-empty compact set in $\mathbb{C}$ .

$\emph{Remark}$ : For Banach algebras over $\mathbb{R}$ the spectrum of an element is also a compact set, although it can be empty. To assure that it is not the empty set, proofs usually involve Liouville's theorem for functions of a complex variable with values in a Banach algebra.

Proof : Let $e$ be the identity element of $\mathcal{A}$ . Let $\sigma(a)$ denote the spectrum of the element $a \in \mathcal{A}$ .

• Compactness - For each $\lambda \in \mathbb{C}$ such that $|\lambda| > \|a\|$ one has $\|\lambda^{-1}a\|<1$ , and so, by the Neumann series, $e-\lambda^{-1}a$ is invertible. Since
  
  we see that $a-\lambda e$ is also invertible.

  We conclude that $\sigma(a)$ is contained in a disk of radius $\|a\|$ , and therefore it is bounded.

  Let $\phi : \mathbb{C} \longrightarrow \mathcal{A}$ be the function defined by

  

  It is known that the set $\mathcal{G}$ of the invertible elements of $\mathcal{A}$ is open (see this entry).

  Since $\phi^{-1}(\mathcal{G})=\mathbb{C}-\sigma(a)$ and $\phi$ is a continuous function we see that that $\sigma(a)$ is a closed set in $\mathbb{C}$ .

  As $\sigma(a)$ is a bounded closed subset of $\mathbb{C}$ , it is compact.

• Non-emptiness - Suppose that $\sigma(a)$ was empty. Then the resolvent function $R_a$ is defined in $\mathbb{C}$ .

  We can see that $R_a$ is bounded since it is continuous in the closed disk $|\lambda|<\|a\|$ and, for $\lambda >\|a\|$ , we have (again, by the Neumann series) \begin{eqnarray*} \|R_a(\lambda)\| & = & \|(a-\lambda e)^{-1}\| \\ & = & \|\lambda^{-1}(e-\lambda^{-1}a)^{-1}\| \\ & \leq & \frac{|\lambda|^{-1}}{1-|\lambda|^{-1}\|a\|} \\ & = & \frac{1}{|\lambda|-\|a\|} \end{eqnarray*}and therefore $\displaystyle \lim_{|\lambda| \rightarrow \infty} R_a(\lambda) = 0$ , which shows that $R_a$ is bounded.

  The resolvent function, $R_a$ , is analytic (see this entry). As it is defined in $\mathbb{C}$ , it is a bounded entire function. Applying Liouville's theorem we conclude that it must be constant (see this this entry for an idea of how Liouville's theorem holds for Banach space valued functions).

  Since $R_a(\lambda)$ converges to $0$ as $|\lambda| \rightarrow \infty$ we see that $R_a$ must be identically zero.

  Thus, we have arrived to a contradiction since $0$ is not invertible.

  Therefore $\sigma(a)$ is non-empty.$\square$